The vapour pressure of pure liquid solvent is 0.50 atm. When a non-volatile solute B is added to the solvent, its vapour pressure drops to 0.30 atm. Thus, mole fraction of the component B is

Question

The vapour pressure of pure liquid solvent is 0.50 atm. When a non-volatile solute B is added to the solvent, its vapour pressure drops to 0.30 atm. Thus, mole fraction of the component B is (A) 0.6 (B) 0.25 (C) 0.45 (D) 0.75

Tardigrade Admin · Accepted Answer

Binary mixture contains non-volatile solute (B) in volatile solvent
By Raoult's law, relative decrease in vapour pressure =

\frac{Δ P}{P ^{o}} = χ_{Solute}

\Rightarrow \frac{0.30}{0.50} = χ_{Solute}

∴ χ_{Solute} = 0.6

Q. The vapour pressure of pure liquid solvent is $0.50 a t m$ . When a non-volatile solute $B$ is added to the solvent, its vapour pressure drops to $0.30 a t m$ . Thus, mole fraction of the component $B$ is

$0.6$

$0.25$

$0.45$

$0.75$

Binary mixture contains non-volatile solute (B) in volatile solvent
By Raoult's law, relative decrease in vapour pressure =
$\frac{Δ P}{P ^{o}} = χ_{Solute}$
$\Rightarrow \frac{0.30}{0.50} = χ_{Solute}$
$∴ χ_{Solute} = 0.6$

Q. The vapour pressure of pure liquid solvent is 0.50atm. When a non-volatile solute B is added to the solvent, its vapour pressure drops to 0.30atm. Thus, mole fraction of the component B is

0.6

0.25

0.45

0.75